The equation of a circle $C$ is $x^2+y^2+10x+12y+52 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2+10x) + (y^2+12y) = -52$ $(x^2+10x+25) + (y^2+12y+36) = -52 + 25 + 36$ $(x+5)^{2} + (y+6)^{2} = 9 = 3^2$ Thus, $(h, k) = (-5, -6)$ and $r = 3$.